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3 years ago

An object moving with a speed of 5m/s has a kinetic energy of 100J what is the mass of the object

Mathematics

1 answer:

Sphinxa [80]3 years ago

5 0

Answer:

k. e. = 1/2 mv^2

100 = 1/2 * m * 5^2

100 = 1/2 * m * 25

m = 100* 2/25

m = 8 kg

hope it helps you

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PLEASE HELP I GIVE BRAINLIEST!! Find the value of x

yawa3891 [41]

Answer: x = 24

Step-by-step explanation:

There are two triangles here; all angles in a triangle add up to 180, so in the first triangle we have two angles.

We add 94 and 41 together, and this gives us 135.

Now, we find the third angle of this triangle and the angle of the inner triangle by doing 180 - 135.

This gives us 45, and now we can calculate for x using the same method we used above.

111 + 45 = 156

180 - 156 = 24

Thus, x is equal to 24 degrees.

4 0

3 years ago

Will someone please help me understand this?

mario62 [17]

Option A:

The length of diagonal JL is $3 \sqrt{5} \text { units }$ .

Solution:

In the quadrilateral, the coordinates of J is (1, 6) and L is (7, 3).

So that, $x_1=1, y_1=6, x_2=7, y_2=3$

To find the length of the diagonal JL.

Using distance formula:

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

$d=\sqrt{(3-6)^2+(7-1)^2}$

$d=\sqrt{(-3)^2+(6)^2}$

$d=\sqrt{9+36}$

$d=\sqrt{45}$

$d=\sqrt{3^2\times 5}$

$d=3\sqrt{ 5}$ units

The length of diagonal JL is $3 \sqrt{5} \text { units }$ .

Option A is the correct answer.

7 0

3 years ago

Read 2 more answers

The sum of the series {1(2/3)}²+{2(1)3)}²+3²+{3(2/3)}²+....to 10 term is

ryzh [129]

Step-by-step explanation:

<h3><u>Given Question </u></h3>

The sum of the series is

$\tt{ {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - 10 \: terms}$

$\green{\begin{gathered}\large{\sf{{\underline{Formula \: Used - }}}} \end{gathered}}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n}1 = n \: }}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n}k = \frac{n(n + 1)}{2} \: }}$

$\boxed{\tt{ \displaystyle\sum_{k=1}^{n} {k}^{2} = \frac{n(n + 1)(2n + 1)}{6} \: }}$

$\large\underline{\sf{Solution-}}$

Given series is

$\rm :\longmapsto\: {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - - 10 \: terms$

can be rewritten as

$\rm \: = \: {\bigg[\dfrac{5}{3} \bigg]}^{2} + {\bigg[\dfrac{7}{3} \bigg]}^{2} + {\bigg[\dfrac{9}{3} \bigg]}^{2} + {\bigg[\dfrac{11}{3} \bigg]}^{2} + - - - 10 \: terms$

$\rm \: = \: \dfrac{1}{9}[ {5}^{2} + {7}^{2} + {9}^{2} + - - - 10 \: terms \: ]$

Now, here, 5, 7, 9 forms an AP series with first term 5 and common difference 2.

So, its general term is given by 5 + ( n - 1 )2 = 5 + 2n - 2 = 2n + 3

So, above series can be represented as

$\rm \: = \: \dfrac{1}{9}\displaystyle\sum_{n=1}^{10}(2n + 3) ^{2}$

$\rm \: = \: \dfrac{1}{9}\displaystyle\sum_{n=1}^{10}\bigg[ {4n}^{2} + 9 + 12n\bigg]$

$\rm \: = \: \dfrac{1}{9}\bigg[\displaystyle\sum_{n=1}^{10} {4n}^{2} + \displaystyle\sum_{n=1}^{10}9 + 12\displaystyle\sum_{n=1}^{10}n\bigg]$

$\rm \: = \: \dfrac{1}{9}\bigg[4\displaystyle\sum_{n=1}^{10} {n}^{2} +9 \displaystyle\sum_{n=1}^{10}1 + 12\displaystyle\sum_{n=1}^{10}n\bigg]$

$\rm \: = \: \dfrac{4}{9}\bigg[\dfrac{10(10 + 1)(20 + 1)}{6} \bigg] + 10 + \dfrac{4}{3}\bigg[\dfrac{10(10 + 1)}{2} \bigg]$

$\rm \: = \: \dfrac{4}{9}\bigg[\dfrac{10(11)(21)}{6} \bigg] + 10 + \dfrac{4}{3}\bigg[\dfrac{10(11)}{2} \bigg]$

$\rm \: = \: \dfrac{1540}{9} + 10 + \dfrac{220}{3}$

$\rm \: = \: \dfrac{1540 + 90 + 660}{9}$

$\rm \: = \: \dfrac{2290}{9}$

Hence,

$\boxed{\tt{ {\bigg[1\dfrac{2}{3} \bigg]}^{2} + {\bigg[2\dfrac{1}{3} \bigg]}^{2} + {3}^{2} + {\bigg[3\dfrac{2}{3} \bigg]}^{2} + - - 10 \: terms = \frac{2290}{9}}}$